LeetCode in Kotlin
1325. Delete Leaves With a Given Value
Medium
Given a binary tree root and an integer target, delete all the leaf nodes with value target.
Note that once you delete a leaf node with value target, if its parent node becomes a leaf node and has the value target, it should also be deleted (you need to continue doing that until you cannot).
Example 1:
Input: root = [1,2,3,2,null,2,4], target = 2
Output: [1,null,3,null,4]
Explanation: Leaf nodes in green with value (target = 2) are removed (Picture in left). After removing, new nodes become leaf nodes with value (target = 2) (Picture in center).
Example 2:
Input: root = [1,3,3,3,2], target = 3
Output: [1,3,null,null,2]
Example 3:
Input: root = [1,2,null,2,null,2], target = 2
Output: [1]
Explanation: Leaf nodes in green with value (target = 2) are removed at each step.
Constraints:
- The number of nodes in the tree is in the range
[1, 3000]. 1 <= Node.val, target <= 1000
Solution
import com_github_leetcode.TreeNode
/*
* Example:
* var ti = TreeNode(5)
* var v = ti.`val`
* Definition for a binary tree node.
* class TreeNode(var `val`: Int) {
* var left: TreeNode? = null
* var right: TreeNode? = null
* }
*/
@Suppress("NAME_SHADOWING")
class Solution {
fun removeLeafNodes(root: TreeNode?, target: Int): TreeNode? {
var root = root
while (hasTargetLeafNodes(root, target)) {
root = removeLeafNodes(target, root)
}
return root
}
private fun removeLeafNodes(target: Int, root: TreeNode?): TreeNode? {
var root = root
if (root == null) {
return root
}
if (root.`val` == target && root.left == null && root.right == null) {
root = null
return root
}
if (root.left != null && root.left!!.`val` == target && root.left!!.left == null && root.left!!.right == null) {
root.left = null
}
if (root.right != null && root.right!!.`val` == target && root.right!!.left == null &&
root.right!!.right == null
) {
root.right = null
}
removeLeafNodes(target, root.left)
removeLeafNodes(target, root.right)
return root
}
private fun hasTargetLeafNodes(root: TreeNode?, target: Int): Boolean {
if (root == null) {
return false
}
return if (root.left == null && root.right == null && root.`val` == target) {
true
} else hasTargetLeafNodes(root.left, target) || hasTargetLeafNodes(root.right, target)
}
}